If two distributions vanish on the same set of test functions, then one is a constant multiple of the other

Question

From Chapter 1, Exercise 14, Strichartz's book: Distribution theory & fourier transforms

14. Suppose $f$ & $g$ are distributions such that $\langle f,\phi\rangle=0 \Leftrightarrow \langle g,\phi \rangle = 0$. Show that $\langle f,\phi \rangle =c\langle g,\phi\rangle$ for some constant c.

Does the question mean that if the condition is true for some test functions $\phi$ the result $\langle f,\phi \rangle=c \langle g,\phi\rangle$ can be deduced for all such test functions $\phi$? I think that one has to exploit the continuity of distributions like one has for functions wherein continuous functions are completely defined if they are defined on a dense subset. But apart from that I don't know how to proceed.

Answer 1

The assumption: $\forall \phi $ (in the test class) $\langle f,\phi\rangle=0 \Leftrightarrow \langle g,\phi \rangle = 0$.

The desired conclusion: $\exists c$ such that $\forall \phi$ the equality $\langle f,\phi \rangle =c\langle g,\phi\rangle$ holds.

Plan of attack:

1. Dispose of the trivial case in which $\forall \phi $ $\langle g,\phi\rangle=0$.
2. Pick $\phi_0$ such that $\langle g,\phi_0\rangle\ne 0$. Let $c=\langle f,\phi_0 \rangle /\langle g,\phi_0\rangle$; this is the only choice that could work, right?
3. If $\phi$ is any test function, let $b=\langle g,\phi \rangle /\langle g,\phi_0\rangle$ and observe that $\langle g,\phi-b\phi_0\rangle=0$. The rest should be easy.